Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is an extension to the question I asked earlier here. The overarching question was the following homework problem,

Let $X_{n} \rightarrow X$ in probability. Show that $\liminf_{n} \mathbb{E}[X_{n}] \ge \mathbb{E}[X]$.

Obviously, this is very similar to the statement of Fatou's Lemma. I also know the following,

$X_{n} \rightarrow X$ in probability if and only if for all subsequences $X_{n(m)}$ of $X_{n}$ there exists a sub-subsequence $X_{n(m_{k})} \rightarrow X$ almost surely.

Given what was answered in my last question, it is trivial to show that Fatou's lemma applies to all of these sub-subsequences $X_{n(m_{k})}$, but how can I bring this back to $X_{n}$ itself, being that it lacks almost sure convergence to $X$?

EDIT Another condition is that $X_n \ge 0$

share|cite|improve this question
Missing hypotheses? As stated, this is false. For a counterexample, assume that $P(X_n=0)=1-p_n$ and $P(X_n=x_n)=p_n$ for given $x_n$ and $p_n$. Then $X_n\to X=0$ in probability as soon as $p_n\to0$, but $E(X)=0$ and $E(X_n)=x_np_n$ can be anything one wants. – Did Sep 13 '11 at 19:00
You're right, I forgot to add that all Random Variables must be non negative. – duckworthd Sep 14 '11 at 2:59
up vote 4 down vote accepted

Edit: My first version was a bit too complicated, so here's a better version:

If $C = \liminf\limits_{n\to\infty}\;{\mathbb{E}[X_n]} = \infty$ there's nothing to prove, so assume that $C \lt \infty$.

We may choose a subsequence $X_{n(m)}$ such that $\mathbb{E}[X_{n(m)}] \to C = \liminf\limits_{n\to\infty}\;{\mathbb{E}[X_{n}]}$ by the definition of the $\liminf$. As you stated in your question, there's a sub-subsequence $X_{n(m_k)}$ such that $X_{n(m_k)} \to X$ a.e. since $X_n \to X$ in measure. As $X_n \geq 0$ a.e., the pointwise a.e. Fatou lemma gives us $\mathbb{E}[X] \leq \liminf\limits_{k \to \infty}\;E[X_{n(m_k)}]= \liminf\limits_{n\to\infty}\;{\mathbb{E}[X_n]}$, as desired.

share|cite|improve this answer
What do you mean by $\mathbb{E}[X_{n(m)}] < C + \epsilon$? Should that be $\lim_{m \rightarrow \infty} \mathbb{E}[X_{n(m)}] < C + \epsilon$? I think the proof goes through if that was your intent. – duckworthd Sep 14 '11 at 17:58
I meant what I wrote. Note that $\mathbb{E}[X_{n}]$ is a sequence of real numbers that accumulates to $C$, thus for every $\varepsilon \gt 0$ there must be infinitely many $X_n$'s such that $\mathbb{E}[X_n]$ is smaller than $C + \varepsilon$. Just pick your subsequence among those. But you can of course also do what you suggest. – t.b. Sep 14 '11 at 18:10
@duckworthd: I've updated my answer following your suggestion. The previous version was correct but a bit too complicated, sorry about that. I hope that it's fine now. – t.b. Sep 15 '11 at 2:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.